Yahoo! 3 years ago (cos A + cos b)^2 + (sin A + sin B)^2=?

1. Yahoo!

@apoorvk @satellite73 @A.Avinash_Goutham @ajprincess @AndrewNJ

2. apoorvk

okay, expand the two whole-squares, it's simple enough...

3. Yahoo!

the options are 4cos^2(a-b/2) 4 sin a/2 cos b/2

4. Yahoo!

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5. Yahoo!

@apoorvk after expanding i got this then next..............

6. apoorvk

great, now.. cosAcosB + sinAsinB = cos(A-B) Use this^

7. patmyer1

ugh, trig identities are a drag. Even after deriving them myself I simply CANNOT seem to memorize them

8. AndrewNJ

spread it so we have (cosa)^2+(cosb)^2+2(cosa)(cosb)+(sina)^2+(sinb)^2+2(sina)(sinb) =2+2cos(a-b)

9. AndrewNJ

(cosa)^2+(sina)^2=1 (cosb)^2+(sinb)^2=1 (cosa)(cosb)+(sina)(sinb)=cos(a-b)

10. Yahoo!

ok @apoorvk so it is not over yet!!@

11. shivam_bhalla

Continuing from where @AndrewNJ left, $2(1+\cos(a-b))$ now we know that $1 + \cos 2a = 2 \cos^2 a$ Now try :D

12. Yahoo!

@shivam_bhalla did nt understand!

13. shivam_bhalla

= $2(1+\cos 2\frac{(a-b)}{2})$ Now we have a identity, $1 + \cos 2x = 2 \cos^2 x$ in your case, $x = \frac{a-b}{2}$ Now try

14. Yahoo!

i got it thxx @shivam_bhalla

15. shivam_bhalla